Unit 8 · Topic 8.4

8.4 Electric Fields of Charge Distributions

When charge is spread along a rod, ring or arc, you can't use the point-charge formula once. Instead you split the charge into tiny pieces, add up their fields with an integral, and let symmetry cancel the components that don't survive. This is the calculus heart of Unit 8.

Key terms

continuous charge distribution
charge element dq
linear charge density (λ)
surface charge density (σ)
volume charge density (ρ)
symmetry

Charge densities

A continuous charge distribution is charge spread smoothly over a line, surface or volume. You describe it with a charge density:

Linear charge density λ (C/m), for a rod or ring: $dq = \lambda\,dx$ or $dq = \lambda\,ds$ .
Surface charge density σ (C/m²), for a sheet or plate: $dq = \sigma\,dA$ .
Volume charge density ρ (C/m³), for a solid: $dq = \rho\,dV$ .

The method

1. Draw the distribution and pick a small charge element dq at a general position. Write dq in terms of the density and a length, like $dq = \lambda\,dx$ . If λ varies (say $\lambda = \alpha x$ ), put that function in.

2. Write the field of that element as if it were a point charge: $dE = \dfrac{1}{4\pi\varepsilon_0}\dfrac{dq}{r^2}$ , where r is the distance from dq to the point.

3. Use symmetry. For each dq, find its mirror-image partner. Components that the partner cancels add to zero, so only integrate the component that survives (often with a factor like $\cos\theta$ or $\sin\theta$ ).

4. Integrate over the whole distribution, keeping track of which quantities vary (x, r, θ) and which are constant.

5. Check limits. Far away, any finite distribution should look like a point charge: $E \to \dfrac{kQ}{r^2}$ .

Results you should be able to derive

The exam asks for integration only for the shapes in this table. With total charge Q:

Distribution	Field
Ring of radius R, on its axis at distance x	$E = \dfrac{kQx}{(x^2 + R^2)^{3/2}}$ , along the axis
Semicircular arc of radius R, at its center	$E = \dfrac{2k\lambda}{R} = \dfrac{2kQ}{\pi R^2}$
Rod of length L, on its own line, distance d from the near end	$E = \dfrac{kQ}{d(d + L)}$
Rod of length L, on its perpendicular bisector, distance y away	$E = \dfrac{kQ}{y\sqrt{y^2 + L^2/4}}$
Infinite line or cylinder, distance r from the axis	$E = \dfrac{2k\lambda}{r} = \dfrac{\lambda}{2\pi\varepsilon_0 r}$

Reading the results

The ring's field is zero at its center, where every piece is canceled by the piece across from it. It rises to a maximum and then falls off as kQ/x² far away. The infinite line falls off only as 1/r, more slowly than a point charge, because there's always more charge farther along the line. The infinite line is much easier with Gauss's law (8.6), which is what most students use.

Part of a semicircle, like a quarter circle, works the same way: integrate θ over just that piece. With less symmetry, both components may survive, so integrate each one. You won't be asked to integrate for a full disk or other shapes outside the table, so don't spend your study time there.

Worked examples

Try each one yourself first, then open the solution.

Example 1Calculator allowed
Rod on its own line
A thin rod of length L carries charge Q spread uniformly. Point P is on the line of the rod, a distance d beyond one end. Derive the field at P.
Show the solutionHide the solution
1. Step 1: Put P at the origin and let the rod run from x = d to x = d + L. Then $\lambda = Q/L$ and a piece at x has $dq = \lambda\,dx$ .
2. Step 2: Every piece is on the same line, so every dE points the same way (away from the rod if Q is positive). No components cancel, and $dE = \dfrac{k\lambda\,dx}{x^2}$ .
3. Step 3: Integrate: $E = k\lambda\displaystyle\int_d^{d+L}\frac{dx}{x^2} = k\lambda\left(\frac{1}{d} - \frac{1}{d+L}\right) = \frac{k\lambda L}{d(d+L)}$ .
4. Step 4: With $\lambda L = Q$ : $E = \dfrac{kQ}{d(d+L)}$ . Check: when d is much larger than L, this becomes kQ/d², a point charge.
Answer: $E = \dfrac{kQ}{d(d+L)}$ , directed along the rod's line, away from the rod for positive Q
Example 2Calculator allowed
Ring on its axis
A ring of radius 0.10 m carries +5.0 nC spread uniformly. Find the field on its axis 0.10 m from the center.
Show the solutionHide the solution
1. Step 1: Each piece dq is the same distance $r = \sqrt{x^2 + R^2}$ from the point. The components perpendicular to the axis cancel in pairs from opposite sides of the ring.
2. Step 2: The axial part of each piece is $dE\cos\theta$ , with $\cos\theta = x/r$ . Since x and r are the same for every piece, $E = \dfrac{kx}{r^3}\displaystyle\int dq = \dfrac{kQx}{(x^2+R^2)^{3/2}}$ .
3. Step 3: Numbers: $E = \dfrac{(9.0\times10^{9})(5.0\times10^{-9})(0.10)}{(0.020)^{3/2}} \approx 1.6\times10^{3}$ N/C.
Answer: About 1.6 × 10³ N/C, along the axis, away from the ring
Example 3Calculator allowed
Semicircular arc (classic trap)
A thin rod bent into a semicircle of radius 0.10 m carries +4.0 nC spread uniformly. Find the field at the center of the semicircle.
Show the solutionHide the solution
1. Step 1: Every piece is distance R from the center, so $dE = \dfrac{k\lambda\,ds}{R^2}$ with $ds = R\,d\theta$ and $\lambda = \dfrac{Q}{\pi R}$ .
2. Step 2: Put the arc above the center, from θ = 0 to π. The horizontal components cancel in pairs. The vertical components add: $E = \displaystyle\int_0^{\pi}\frac{k\lambda\sin\theta}{R}\,d\theta = \frac{2k\lambda}{R}$ .
3. Step 3: Substitute λ: $E = \dfrac{2kQ}{\pi R^2} = \dfrac{2(9.0\times10^{9})(4.0\times10^{-9})}{\pi(0.10)^2} \approx 2.3\times10^{3}$ N/C, pointing away from the arc.
4. Step 4: The trap is adding magnitudes without the $\sin\theta$ . That gives kQ/R² = 3.6 × 10³ N/C, too big by a factor of π/2.
Answer: About 2.3 × 10³ N/C, pointing from the arc's middle through the center and beyond

Common mistakes

Integrating the full magnitude dE when symmetry says only one component survives. Multiply by the right $\cos\theta$ or $\sin\theta$ first.
Treating r as constant when it changes from piece to piece (it's constant for a ring or arc, but not for a rod).
Writing dq = λ dθ for an arc. The length of a small piece of arc is R dθ, so dq = λR dθ.
Forgetting to check limits. If your answer doesn't become kQ/r² far away, something is wrong.

On the exam

Free-response questions often ask you to set up the integral, including limits, before solving it. Points come from a correct dq, a correct distance and the right component, so show each one.
You may get a nonuniform density such as λ = αx. Find the total charge by integrating λ dx; don't multiply λ by L.
Multiple-choice questions often test the limits: which expression becomes kQ/x² far away, or which is zero at the center of a ring.

Connected topics

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Check yourself

4 questions on 8.4 Electric Fields of Charge Distributions. Pick an answer to see if you got it, and why.

Question 1 of 4Calculator allowed

A very long, thin rod carries a uniform charge of 3.0 nC per meter. What is the magnitude of the electric field 0.15 m from the rod, away from its ends?

Question 2 of 4Calculator allowed

A thin ring of radius 0.10 m carries 5.0 nC spread evenly around it. What is the field on the ring's axis 0.10 m from its center? (On the axis, $E = \frac{kQx}{(x^2 + R^2)^{3/2}}$ .)

Question 3 of 4Calculator allowed

A ring of radius R carries charge Q. On its axis, $E(x) = \frac{kQx}{(x^2 + R^2)^{3/2}}$ . At what distance x from the center is the field strongest?

Question 4 of 4Calculator allowed

A thin rod bent into a semicircle of radius R carries charge Q spread evenly along it. What is the magnitude of the electric field at the center of the semicircle? ( $k = \frac{1}{4\pi\varepsilon_0}$ )

0 of 4 answered